a: Giá mới của chiếc tivi là:

\(8500000\left(1-15\%\right)=8500000\cdot0,85=7225000\left(đồng\right)\)

b: Giá gốc trước khi giảm là:

\(8000000:\left(1-25\%\right)=8000000:\dfrac{3}{4}=\dfrac{32000000}{3}\left(đồng\right)\)

a: Giá mới của cuốn sách là:

\(48000\left(1-15\%\right)=48000\cdot0,85=40800\left(đồng\right)\)

b: giá mới của cuốn sách là:

\(48000\left(1+10\%\right)=52800\left(đồng\right)\)

m>n

\(A=1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+8}\)

\(=1+\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{8\cdot\dfrac{9}{2}}\)

\(=\dfrac{2}{2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{8\cdot9}\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=2\left(1-\dfrac{1}{9}\right)=2\cdot\dfrac{8}{9}=\dfrac{16}{9}\)

mn nhanh giùm em ạ

ĐKXĐ: x<>1

Để A là số nguyên thì \(5⋮x-1\)

=>\(x-1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{2;0;6;-4\right\}\)

\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{47\cdot49}\\ A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\\ A=\dfrac{1}{1}-\dfrac{1}{49}\\ A=\dfrac{48}{49}\)

Vậy \(A=\dfrac{48}{49}\)

\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{47\cdot49}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{47.49}\)

\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\)

\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{49}\)

\(\Rightarrow2A=\dfrac{49}{49}-\dfrac{1}{49}\)

\(\Rightarrow2A=\dfrac{48}{49}\)

\(\Rightarrow A=\dfrac{48}{49}:2.\)

\(\Rightarrow A=\dfrac{48}{49}\cdot\dfrac{1}{2}\)

\(\Rightarrow A=\dfrac{24}{49}\).

-20/23 + 2/3 - 3/23 + 2/5

= (-20/23 - 3/23) + (2/3 + 3/5)

= -1 + 19/15

= 4/15

4/3 + -11/31 + 3/10 - 20/31

= (4/3 + 3/10) + (-11/31 - 20/31)

= 49/30 + -1

= 19/30

916.7% nha

10≤n≤99↔21≤2n+1≤201

2n+1 là số chính phương nên

2n + 1 ∈{25;49;81;121;169}

=>n ∈ {12;24;40;60;84}

=> 3n + 1 ∈{37;73;121;181;253}

=> n = 40